Scalar Vector Multiplication And Linear Combinations Of Vectors

Go back to  'Vectors and 3-D Geometry'


Intuitively, we can expect that if we multiply a vector \(\vec a\) by some scalar \(\lambda \), the support of the vector will not change; only its magnitude and / or its sense will. Specifically, if \(\lambda \) is positive, the vector will have the same direction; only its length will get scaled according to the magnitude of \(\lambda \). If \(\lambda \) is negative, the direction of the product vector will be opposite to that of the original vector; the length of the product vector will depend on the magnitude of \(\lambda \) .

Note that for any vector \(\vec a\), if we denote the unit vector along \(\vec a\) by \(\hat a\), we have

\[\boxed{\vec a = \left| {\,\vec a\,} \right|\hat a}\]

Put in words, if we multiply the unit vector along a vector  \(\vec a\) by its magnitude, we obtain that vector itself. Put in a slightly different way, we have

\[\hat a = \frac{{\vec a}}{{\left| {\vec a} \right|}}\]

i.e, if we divide a vector by its magnitude, we obtain the unit vector along that vector’s direction.

Another very important result that follows from this discussion is that two vectors \(\vec a\,\,{\text{and}}\,\,\vec b\) are collinear if and only if there exists some  \(\lambda  \in \mathbb{R}\) such that

\[\boxed{\vec a = \lambda \vec b}{\quad \mathbf{Collinear}}{\text{ }}{\mathbf{vectors}}\]

i.e, two vectors are collinear if one can be obtained from the other simply by multiplying the latter with a scalar.

This fact can be stated in another way : consider two non-collinear vectors \(\vec a\,\,{\text{and}}\,\,\vec b\)  . If for some \(\lambda ,\mu  \in \mathbb{R}\), the relation

\[\lambda \vec a + \mu \vec b = \vec 0\quad\quad\quad...{\text{ }}\left( 1 \right)\]

is satisfied, then \(\lambda \,and\,\mu \)  must be zero. This is because (1) can be written as

\[\vec a = \left( { - \frac{\mu }{\lambda }} \right)\vec b\]

which would imply that \(\vec a\)  is a scalar multiple of \(\vec b\) ,i.e., \(\vec a\,\,{\text{and}}\,\,\vec b\) are collinear, contradicting our initial supposition that \(\vec a\,\,{\text{and}}\,\,\vec b\) are non-collinear.

In subsequent discussions, we’ll be talking a lot about linear combinations of vectors. Let us see what we mean by this. Consider n arbitrary vectors \({\vec a_1},{\vec a_2}......{\vec a_n}.\) A linear combination of these n vectors is a vector \(\vec r\) such that

\[\vec r = {\lambda _1}{\vec a_1} + {\lambda _2}{\vec a_2} + ...... + {\lambda _n}{\vec a_n}\quad\quad\quad...{\text{ }}\left( 2 \right)\]

where \({\lambda _1},{\lambda _2}....{\lambda _n} \in \mathbb{R}\) are arbitrary scalars. Any sort of combination of the form in (2) will be termed a linear combination.

Thus, using the terminology of linear combinations, we can restate the result we obtained earlier: for any two non-zero and non-collinear vectors  \(\vec a\,\,{\text{and}}\,\,\vec b\), if their linear combination is zero, then both the scalars in the linear combination must be zero.

We now come to a very important concept.

Learn math from the experts and clarify doubts instantly

  • Instant doubt clearing (live one on one)
  • Learn from India’s best math teachers
  • Completely personalized curriculum